PHP - Where Should Data First Be Stored

 wrote a stored procedure this morning and i don’t know how to get the values out of it through a class function in php or phpmyadmin.

here is what i wrote :

public function totalProcedures($friend_name,$session_id) 
{ /* *query to fetch stored procedure */ 
try { 
//executing the stored procedure 
$sql_sp="CALL timeline (:friend, :session,@updates, @group_posts)";
$stmt_sp= $this->_db->prepare($sql_sp); 
$stmt_sp->bindValue(":friend",$friend_name);
$stmt_sp->bindValue(":session",$session_id); 
$stmt_sp->execute();
$rows=$stmt_sp->fetch(PDO::FETCH_ASSOC); 
$stmt_sp->closeCursor(); // closing the stored procedure 
//trying to get values from OUT parameters.
$stmt_sp_2=$this->_db->prepare("select @updates,@group_posts");
$stmt_sp_2->execute();
return $stmt_sp_2->fetch(PDO::FETCH_ASSOC);
} catch (PDOException $ei)
{
echo $ei->getMessage(); } 
}

can someone helpme how to get results.

here is the storedprocedu

DELIMITER $$
CREATE DEFINER=`root`@`localhost` PROCEDURE `timeline`(IN `friend` VARCHAR(255), IN `session_id` VARCHAR(255), OUT `updates` VARCHAR(62555), OUT `group_posts` VARCHAR(62555))
BEGIN


select * FROM updates where author in (friend,session_id)  order by time desc limit 5;



select * FROM group_posts where author_gp in (friend,session_id) order by pdate desc limit 5;
END$$
DELIMITER ;

i get the result in php myadmin as follows:

 how do i do this inside a php class function.

CALL timeline('shan2batman','aboutthecreator', @updates, @group_posts);

 

The path to a image is stored in a variable and I need to display this image in the email. Can someone help me.

This is my current attempt and nothing is happening. But the variable passes the URL correctly.

Code: [Select]
<TABLE BORDER='0'>
<TR>
<TD align='left' width='370'><img src='".$dis_logoimg."' width='370' height='86' /></TD>
</TR>
</TABLE>

I'm working on a project and cannot figure out why one piece of code works in one project but not in this one.

Code: [Select]
$username = "voltageme5";
$result = $mysqli->query("call usernameCheck($username)");

returns
string(43) "Unknown column 'voltageme5' in 'field list'"

Here is the stored routine:

Code: [Select]
CREATE PROCEDURE `database`.`table` (IN memUsername varchar(45))
BEGIN
select count(*) as total from members where username = memUsername;
END

Whats weird is that in the PHP code if i replace the variable with a string, it works. It's only with the variable in the call that it errors out like that. I'm new to this stored procedure thing so I'm sure I'm missing something stupid.

Hi Everyone,

I have this php page that runs couple of MS SQL stored procedures to fetch me values from SQL Server, however I have run into a problem where not more then 2 stored procedures are executed at same time. For instance below mentioned code only fetches me values for 1st and 2nd SP's. If I need values from the 3rd stored procedure then I would have to comment either one of the first two stored procedure.   Can someone please tell me what is that I am doing wrong.   I am relatively new to PHP, so please be gentle. Any help would be highly appreciated.

Thanks.

<?php


if($conn = mssql_connect('host', 'user', 'pass')) echo 'Connected to SQLEXPRESS<BR>';
if(mssql_select_db("database",$conn)) echo 'Selected DB: SomeDatabase<BR>';

$variable1=something;

$sql_statement =  mssql_init("stored_procedure1 '$variable1'", $conn);
$result=mssql_execute($sql_statement);

$x=0;

while ($row = mssql_fetch_assoc($result))
    {
    $column2[$x]= $row['column2'];
    $column1= $row['column1'];
    
$x++; }

    
echo "Value 1: $column1      </br>";
echo "Value 2: $column2[0]  </br>";
echo "Value 3: $column2[1]  </br>";
echo "Value 4: $column2[2]  </br>";
echo "Value 5: $column2[3]  </br>";
echo "Value 6: $column2[4]  </br>";
echo "Value 7: $column2[5]  </br>";


mssql_free_statement($sql_statement);



$sql_statement =  mssql_init("stored_procedure2 '$variable1'", $conn);
$result=mssql_execute($sql_statement);


$x=0;

while ($row = mssql_fetch_assoc($result))
    {
    $someval1[$x]= $row['somecolumn1'];
    $someval2[$x]= $row['somecolumn2'];
    $someval3= $row['somecolumn3'];
    $someval4= $row['somecolumn4'];
    $someval5= $row['somecolumn5'];
    $someval6= $row['somecolumn6'];

$x++; }

    
echo "Something1: $someval1[0]</br>";
echo "Something2: $someval3 </br>";
echo "Something3: $someval2[0] </br>";
echo "Something4: $someval4 </br>";
echo "Something5: $someval6 </br>";
echo "Something6: $someval5 </br>";


mssql_free_statement($sql_statement);


$sql_statement =  mssql_init("stored_procedure3 '$variable1'", $conn);
$result=mssql_execute($sql_statement);


$x=0;

while ($row = mssql_fetch_assoc($result))
    {
    $somevaluecheck= $row['somecolumncheck'];

$x++; }
    
echo "SomethingCheck: $somevaluecheck </br>";


mssql_free_statement($sql_statement);


?>

Hey guys,

I have been banging my head against a wall here with this. I am saving my sessions in my database via session_set_save_handler(). So let me walk you through what I have here, what works, and what the issue seems to be.

basically, the problem is the $_SESSION array is empty on page load.

common.php:
I have the open / close / read / write / destroy / and gc functions. These all work properly as when i use a session variable it stores into the database and i can see all of the information in there..

inside of common.php i have session_start().. I am positive that the session_start() is running becasue common.php is included into index.php and i tried adding session_start() to index.php again and i got an E_NOTICE saying the session already began.

(yes, for the read function i am returning a string... i included that below).

for the table itself, i have set the session_data as both text and blob, same issue with both.


index.php:
includes common.php. I know it's included as other aspects of the file are included and work properly.

if i call var_dump($_SESSION) i get an empty array.  and i prited out the session_id() and it matched my session id in the database. and the values that are stored in there are correct. i have for example:

count|i:0

now with that value actually in the database and when calling session_id() and i get the ID that matches in the table. so i will get an E_NOTICE of an undefined index..

i was trying something like:


if(!isset($_SESSION['count']))
     $_SESSION['count'] = 0;
else
     $_SESSION['count']++;

echo $_SESSION['count'];


Everytime the page is reloaded, count is reset to 0.. I can tell that it is reset to 0 as the expired time changes in the database after every load of the page (which is part of the write function).  I double checked that it wasn't a problem for some reason with the ++ by adding in a variable that sets to 1 when in the if, and 0 if in the else, and it always outputs a 1.

i have included here my read function since that apparently is the issue..

function sess_read($sess_id)
{
      global $DB;      

      $sql = "SELECT `session_data` FROM `sessions` WHERE
                session_id = '".$sess_id."' AND
                session_agent = '".$_SERVER["HTTP_USER_AGENT"]."' AND
                session_expire > '".time()."';";                

       $query = $DB->query($sql);       

       if($DB->num_rows($query) > 0)       
       {

       $r = $DB->fetch($query);       
           return settype($r['session_data'], 'string');           

   }       

   

        
       return '';
}

I tried also with removing the agent and expire check to see if it was an issue there but same problem.

I probably have missed something really dumb but i can't for the life of me figure it out and I have googled around for a similar issue. The actual output of the page is correct.. all of the HTML and CSS information loads properly.. no errors are reported (and i have E_ALL on).

Thanks

I have a PHP page that offers various information from a single text file. This text file is

encrypted on the server HD.

Upon initial entry into the page, the user enters an encryption/decryption KEY and the

encrypted file is decrypted to clear text and it is available for viewing.

I have some parameters that I store in PHP session variables. I do this since various

subsequent actions by the user will require these parameters. The code is written and the whole process seems to

work well.

Since the info in these session variables is sensitive, I need to understand WHERE they are

stored. I know that it is a file on the HD, but after hours of reading the PHP Manual on

sessions, I am not finding where (HD directory) that storage is.

I have a typical shared hosting account for my web site. Mostly I want to discover is, are

the session variables in y User/file hierarchy, or are they stored in a system area where

the PHP is installed.

Whew. Sorry this was so long.

Thank you,
xphp

I am trying to make a  setup file for a website I'm developing, and am having the database tables and stored procedures created automatically.  However, I'm having issue getting the stored procedures to be created.  Here is my code:
Code: [Select]
$db->Q("
DELIMITER $$
CREATE PROCEDURE `Database`.`Procedure_Name`(in sp_uid mediumint(20) unsigned, in sp_user varchar(15), in sp_pass varchar(20), in sp_email varchar(30))
BEGIN
Insert Into Table Values (sp_uid, sp_user, md5(sp_pass), sp_email, '1', '1', '0');
Select true;
END$$
");
Assume that the "$db->Q()" works just fine, as I'm having issues no where else with it.  This automatically connects to the database and runs a query with whatever is inside the "()".  The tables are being created just fine, but no stored procedures are being created.  I've tried everything I can think of, and googled my question many different ways without finding an answer or work-through.  Does anyone know what I am doing wrong?

Thanks in advance.

I am writting a php function that uses mysql to get user data - pretty common, right

Well, my issue is that I need to run a check in my file system.  Users profile pictures are stored in my image directory as .png's.  I need to have my function check that directory and if an image matches their id, then return their information.  I only want the user data if they have an image uploaded.

Here is my current function:

Code: [Select]
function fetch_users_login($limit)
{
$limit = $limit(int);

$sql = "SELECT
                `users`.`id`,
                `users`.`firstname`,
                `users`.`lastname`,
                `users`.`username`,
`user_privacy`.`avatar`
            FROM `users`
LEFT JOIN `user_privacy`
            ON `users`.`id` = `user_privacy`.`uid`
WHERE `users`.`status` > 2
AND `user_privacy`.`avatar` = 1
ORDER BY `users`.`id` DESC
LIMIT 0, {$limit}";

$result = mysql_query($sql) or die(mysql_error());

$users = array();

$i = 0;

while(($row = mysql_fetch_assoc($result)) !== false)
{
$users[$i] = array(
'id' => $row['id'],
'firstname' => $row['firstname'],
'lastname' => $row['lastname'],
);

$users[$i]['avatar'] = getUserAvatar($row['username']);

$i++;
}

return $users;
}

Hello,

Here's my system. Once a user is successfully logged in, a new instance of a User class is created. The constructor of this class grabs the details of the logged-in user from a database and stores them inside properties in the User class.

The problem is, obviously I'd want to access these properties from any page I require them to be able to display user data, so I looked into storing the User object in a Session.

When I tried implementing this, I ran into a bunch of errors and I couldn't figure it out.

Here's an example:

After a user has logged in, I had the following code:
Code: [Select]
$_SESSION['user'] = new User($this->username);
I was under the impression that this assigns a user object to a session. But it's not working as I receive this error:

Quote

Notice: Undefined index: user in ../v2/admin/index.php on line 18



Then on the page I want to display the name of the current user logged-in, I had this code:
Code: [Select]
$_SESSION['user']->get_Name();
But then I get this error:
Quote

Fatal error: Call to a member function get_IP() on a non-object in ../v2/admin/index.php on line 18



Can tell me what I have to do, to make this work?


Thanks.

How can I run php code that is stored in a database?

I have my pages created by grabbing the contents of a pageBody mySQL table cell, but some pages require further php to be able to display correctly.

For example, I have a players page, which will list all players, and some information about them. The players and information is all saved in a players table in my database (separate to the pages table where pageBody is stored). I use simple php to grab all the players and format all of their information so it can be displayed on the page.

This is the flow of information that I would like:
User clicks on players page browser loads content page (this is a template page), and grabs the pageid from $_GET browser then reads pages table in database to get the pageBody associated with the pageid The pageBody will contain more php, which will run the players script to retrieve the list, and display
Doing it the above way will make it much easier to extend the website to include more types of pages that has to run additional php scripts.
The only way I can think of this working is by doing this:
user clicks on players page
browser loads content page, grabs pageid from $_GET browser checks the pageid for the one associated with players (hard coded into the php script) browser then loads the players.php script instead of going to the database This above way means that I will need to edit the index.php page everytime I add a new list type (example, coaches, volunteers etc).

Anyone have any suggestions? I know my question is long, but I was finding it hard to explain it in an easy to understand way, but I'm still not sure if I have :S.

Cheers
Denno

I really like how Blob and Long blob handle pictures so I decided to try it with audio...I can get the file to upload fine but my problem using the file once its in the database...

This is kind of what I am doing.

This is my file called get_audio.php:

Code: [Select]
<?php
mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("my_db") or die(mysql_error());

$id = addslashes($_REQUEST['id']);

$audio = mysql_query("SELECT* FROM wineAudio_tbl WHERE audio_id=$id");
$audio = mysql_fetch_assoc($audio);
$audio = $audio['audioFile'];

header("Content-type: audio/mp3");

echo $audio;

?>

When I want to use the audio I would just have a link something like this:

Code: [Select]
<a href="get_audio.php?id=12" target="_blank">Play Audio</a>

When you click on the "Play Audio" button it opens a browser mp3 player, but does not start the file..

Any ideas what I could be doing wrong ?

I put together the following blocs of code for uploading pictures into a database and displaying them on a webpage.  The pictures are supposed to be displayed on the member's only page of a website I'm working on, upon logging in, and they are supposed to be the member's uploaded picture.  I created several members and and used one of my existing member accounts to test the uploading process.  The picture upload process appeared to have been successful when I checked on myphpadmin.  Yet,  when I login with this account, no picture is displayed, instead, a tiny jpg icon is displayed at the top left corner of the box in which the picture was supposed to be displayed.  Same thing when I login with the other accounts with which I haven't yet uploaded a picture.

I'll start with the code that installs the table in the database

$query = "CREATE TABLE images (

image_id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY ,
member_id INT UNSIGNED,
like_id INT UNSIGNED,
image LONGBLOB NOT NULL,
image_name varchar(255)  NOT NULL,
image_type varchar(4) NOT NULL,
image_size int(8) NOT NULL,
image_cartegory VARCHAR(20) NOT NULL,
image_path VARCHAR(300),
image_date DATE  

          )";




Then here is the code  which allows the member to upload his pictu
<form enctype="multipart/form-data" action="insert_image.php" method="post" name="changer">
                         <input name="MAX_FILE_SIZE" value="102400" type="hidden">
                         <input name="image" accept="image/jpeg" type="file">
                         <input value="Submit" type="submit">
                        </form>


 And here is the insertimage.php which inserts the image into our database:  Note that I have to authenticate the user in order to register his session id which is used later on in the select query  to identify him and select the right image that corresponds to him.


<?php

//This file inserts the main image into the images table.

//address error handling

ini_set ('display_errors', 1);
error_reporting (E_ALL & ~E_NOTICE);





//authenticate user
//Start session

session_start();

       
        //Connect to database
require ('config.php');

//Check whether the session variable id is present or not. If not, deny access.

if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) {

header("location: access_denied.php");

exit();

}

        else{ 
// Make sure the user actually 
// selected and uploaded a file
if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) { 

      // Temporary file name stored on the server
      $tmpName  = $_FILES['image']['tmp_name'];  
       
      // Read the file 
      $fp      = fopen($tmpName, 'r');
      $data = fread($fp, filesize($tmpName));
      $data = addslashes($data);
      fclose($fp);
      

      // Create the query and insert
      // into our database.
      $query = "INSERT INTO images (member_id, image_cartegory, image_date, image) VALUES ('{$_SESSION['id']}', 'main', NOW(), '$data')";
      $results = mysql_query($query);
      
      // Print results
      print "Thank you, your file has been uploaded.";
      
}
else {
   print "No image selected/uploaded";
}

// Close our MySQL Link
mysql_close();


} //End of if statmemnt.
 


?>  



On the page which is supposed to display the image upon login in, I inserted the following html code in the div that's supposed to contain the image:


                    <div id="image_box" style="float:left; background-color: #c0c0c0; height:150px; width:140px; border-           color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; ">
                 
                         <img src=picscript.php?imname=potwoods>
                             
                    </div>

                   
And finally, the picscript.php  contained the select query:
<?php

                             //address error handling

                             ini_set ('display_errors', 1);
                             error_reporting (E_ALL & ~E_NOTICE);

                             //include the config file
                              require('config.php');

                             $image = stripslashes($_REQUEST[imname]);
                             $rs = mysql_query("SELECT* FROM images WHERE member_id = '".$_SESSION['id']."' AND cartegoty = 'main' ");
                             $row = mysql_fetch_assoc($rs);
                             $imagebytes = $row[image];
                             header("Content-type: image/jpeg");
                             print $imagebytes;

                            
 ?>



Now the million dollar question is, "What is preventing the picture from getting displayed?"  I know this is a very lengthy and laborious problem to follow but I'm sure there is someone out there who can point out where I'm not getting it.  Thanks.

i have made a logon script but he must give a output parameter "relatieid" but if i do print relatieid i don't get the output parameter. what can i do to get the output parameter. i must give -1 if your logon data don't be good and your relatie id if you are succesfull logon. i saw that you can use fatchall but how

my script is:
php
Code: [Select]
<?php 
session_start(); 
$sessieid = session_id(); 
error_reporting(-1); 
ini_set('display_errors', 1); 

if($_SERVER['REQUEST_METHOD'] == 'POST') 
{ 
     $username = $_POST['username']; 
     $wachtwoord = $_POST['wachtwoord']; 
     $ip = $_SERVER["REMOTE_ADDR"]; 
     $computernaam = php_uname('n'); 

     if(trim($username) == '') 
     { 
        echo "<font color='red'>Vul geldige gebruikersnaam in!</font><br>"; 
        header("refresh:5;url=/login/"); 
    exit() ; 
    }     
     
     if(trim($wachtwoord) == '') 
     { 
        echo "<font color='red'>Vul geldige wachtwoord in!</font><br>"; 
        header("refresh:5;url=/login/"); 
    exit() ; 
    } 
      
      
    $db = new PDO('mssql:host=localhost\snelstart;dbname=SluisWWW','test','********'); 
     
    $db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 
     
    $stmt = $db->prepare("EXECUTE spMagInvoeren ?,?,?,?,?,?,?"); 
      
    $stmt->bindValue(1 ,$username, PDO::PARAM_STR); 
    $stmt->bindValue(2 ,$wachtwoord); 
    $stmt->bindValue(3 ,$ip); 
    $stmt->bindValue(4 ,$computernaam); 
    $stmt->bindValue(5 ,$sessieid); 
    $stmt->bindParam(6 ,$poging); 
    $stmt->bindParam(7 ,$relatieid); 

    $stmt->execute(); 

print "<br/>Returned: $relatieid<br/><br/>\r\n"; 





    { 
        setcookie("TestCookie", $username); 
        // redirecten 
        exit(); 
    } 
     
     
} 
stored procedure ms sql server
Code: [Select]
USE [SluisWWW]
GO
/****** Object:  StoredProcedure [dbo].[spMagInvoeren]    Script Date: 04/04/2012 09:54:59 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
-- =============================================
-- Author: Henk Boessenkool
-- Create date: 23 Maart 2012
-- Description: test login
-- =============================================
ALTER PROCEDURE [dbo].[spMagInvoeren]
-- Add the parameters for the stored procedure here
@Usernaam nVarchar(20) ,
@Wachtwoord nvarchar(20),
@IPAdres nvarchar(20),
@Computer nvarchar (20),
@SessieID nvarchar(50),
@PogingenOver integer  OUTPUT,
@RelatieNummer integer OUTPUT
AS
BEGIN
-- SET NOCOUNT ON added to prevent extra result sets from
-- interfering with SELECT statements.
DECLARE @Success INT
SET NOCOUNT ON;
    SET @Success = (SELECT COUNT(*)
        FROM contactpersonen
        WHERE (username = @Usernaam AND wachtwoord = @Wachtwoord))
    IF @Success = 1
    BEGIN
       SET @RelatieNummer = (SELECT TOP 1 relatie_id FROM [SluisWWW].[dbo].[contactpersonen]
                             WHERE (username = @Usernaam AND wachtwoord = @Wachtwoord))
       
    END
    ELSE SET @RelatieNummer = -1
    INSERT INTO [SluisWWW].[dbo].[Logins] (Login,Wachtwoord,RelatieID,IP,Computer,SessieID)
                VALUES (@Usernaam,@Wachtwoord,@RelatieNummer,@IPAdres,@Computer,@SessieID)
               
                set @PogingenOver = 24
 
   
   
   

 
    -- Insert statements for procedure here
END

Hi all,     This is just an experiment to make sure that I can get data from a Stored Proc 'IF' and when I need to, so it's basically a THOUGHT EXERCISE.   MSSQL 2014     I'm a noobie and need a hand, I've managed to get my Instance connected to the internet and I can query it using PHP and SQL, I can also look at views with no problem.     I have it working as an "ADODB.Connection" and like I said it connects and I can query data and display results.     Now I have coded a Stored Proc "GetMonthDays" in Sql Server:   Which returns days 1 through xxx in a given month and also returns the Day name  eg... Sat for each date     2014-01-01 Thurs   2014-01-02 Fri   etc...     It works perfectly and very fast so All cool with that side BUT...     I want to be able to query the Database through a Stored Proc, I've spent all day trying to find a way to get this to work and I've hit a wall   Can anyone please tell me IF it's A. possible and B. maybe if they are really really nice and share with me a code example that I can change to fit my needs ?     This T-SQL returns what it needs to     Begin
  EXEC dbo.qselGetMonthDays '2015-01-01'
  End
      Cheers     JD
Edited by juddadredd, 02 January 2015 - 04:55 PM.

Hi Everyone,

I have a program that generates 200 unique images keeping the first image static in each run.The images keep scrolling on to the screen pause for 3 seconds and scroll off I'm able to generated all 200 unique images without repetition, everything is working well except for the lase two images the last two  images are scrolling on to the screen but are not been displayed in the database, Moreover The last image is a duplicate of 197th image.I don't know what is happening.....

Here is MY code..........



<?php
      session_start();
      $sid = $_SESSION['id'];
      $_SESSION['imageDispCnt'] = 0;
$myQuery = "SELECT * from image";
$conn = mysql_connect("localhost","User","Passwd");
mysql_select_db("database_Name",$conn);
$result = mysql_query($myQuery);
$img =Array();
$id =Array();
$i =0;
$imagepath = 'http://localhost/images/';
while ($row = mysql_fetch_array($result))
      {
      $img[$i] =  $imagepath.$row['img_name'];
      $id[$i] =  $row['imageid'];
      $i = $i + 1;
       }
 ?>
      </head>
      <script language="JavaScript1.2">
var scrollerwidth='800px';
var scrollerheight='600px';
var scrollerbgcolor='white';
var pausebetweenimages=3200;
var s;
var sec;
var d;
var j;
var imgid;
var milisec = 0;
var seconds = 0;
var flag = 1;
var ses_id = '<?php echo $sid;?>';
var count = 0;
var i = 0;
var imgname;
var imgid;
var k =0;
var slideimages=new Array();
var img_id = new Array();
var index;

<?php
  $l =0;   
$count = array();
   $j = rand(0,199);
   while($l < 200)
   {
    while(in_array($j, $count))
        {
           $j = rand(0,199);
        }
        $count[$l] =  $j;
        $l++;
       
     }?>
    <?php
  $k = 0;
  for($k = 0;$k<count($count);$k++){
  ?>
  index = <?php echo $k;?>;
  <?php
        $indx = $count[$k];?>
  if(index == 0){
     slideimages[0] = '<img src="http://localhost/images/hbag044.jpg" name="r_img" id="0"/>';
     img_id[0] = '<input type="hidden" value="0" id="imgId" />';
  }
 
  else if(index > 0)
  {
    slideimages[<?php echo $k?>] = '<img src="<?php echo $img[$indx]?>" name="r_img" id="<?php echo $id[$indx]?>"/>';
     img_id[<?php echo $k?>] = '<input type="hidden" value="<?php echo $id[$indx]?>" id="imgId" />';
  }
<?php
}
?>


Can Any one plese help me

Appreciate your help...


Thanks